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Unformatted text preview: Chapter 3 Section 32 31. Continuous 32. Discrete 33. Continuous 34. Discrete 35. Discrete 36. Continuous 37. Discrete 38. Continuous 39. Continuous Section 33 310. a) Engineers with at least 36 months of fulltime employment. b) Samples of cement blocks with compressive strength of at least 6000 kg per square centimeter. c) Measurements of the diameter of forged pistons that conform to engineering specifications. d) Cholesterol levels at most 180 or at least 220. 311. The intersection of A and B is empty, therefore P(X ∈ A ∩ B) = 0. a) Yes, since P(X ∈ A ∩ B) = 0. b) P(X ∈ A ′ ) = 1  P(X ∈ A) = 1 – 0.4 = 0.6 c) P(X ∈ B ′ ) = 1  P(X ∈ B) = 1 – 0.6 = 0.4 d) P(X ∈ A ∪ B) = P(X ∈ A) + P(X ∈ B)  P(X ∈ A ∩ B) = 0.4 + 0.6 – 0 = 1 312. a) P(X ∈ A ′ ) = 1  P(X ∈ A) = 1 – 0.3 = 0.7 b) P(X ∈ B ′ ) = 1  P(X ∈ B) = 1 – 0.25 = 0.75 c) P(X ∈ C ′ ) = 1  P(X ∈ C) = 1 – 0.6 = 0.4 d) A and B are mutually exclusive if P(X ∈ A ∩ B) = 0. To determine if A and B are mutually exclusive, solve the following for P(X ∈ A ∩ B): P(X ∈ A ∪ B) = P(X ∈ A) + P(X ∈ B)  P(X ∈ A ∩ B) 0.55 = 0.3 + 0.25  P(X ∈ A ∩ B) 0.55 = 0.55  P(X ∈ A ∩ B) and P(X ∈ A ∩ B) = 0. Therefore, A and B are mutually exclusive. e) B and C are mutually exclusive if P(X ∈ B ∩ C) = 0. To determine if B and C are mutually exclusive, solve the following for P(X ∈ B ∩ C): P(X ∈ B ∪ C) = P(X ∈ B) + P(X ∈ C)  P(X ∈ B ∩ C) 0.70 = 0.25 + 0.60  P(X ∈ B ∩ C) 0.70 = 0.85  P(X ∈ B ∩ C) and P(X ∈ B ∩ C) = 0.15. Therefore, B and C are not mutually exclusive. 313. a) P(X > 15) = 1 – P(X ≤ 15) = 1 – 0.3 = 0.7 b) P(X ≤ 24) = P(X ≤ 15) + P(15 < X ≤ 24) = 0.3 + 0.6 = 0.9 c) P(15 < X ≤ 20) = P(X ≤ 20) – P(X ≤ 15) = 0.5 – 0.3 = 0.2 d) P(X ≤ 18) = P(15 < X ≤ 18) + P(X ≤ 15) where P(15 < X ≤ 18) = P(15 < X ≤ 24) – P(18 < X ≤ 24) = 0.6 – 0.4 = 0.2 Therefore, P(X ≤ 18) = P(15 < X ≤ 18) + P(X ≤ 15) = 0.2 + 0.3 = 0.5 Alternatively, P(X ≤ 18) = P(X ≤ 24)  P(18 < X ≤ 24) = 0.9 – 0.4 = 0.5. 314. A  Overfilled, B  Medium filled, C  Underfilled a) P(X ∈ C’) = 1  P(X ∈ C) = 1 – 0.15 = 0.85 b) P(X ∈ A ∪ C) = P(X ∈ A) + P(X ∈ C) – P(X ∈ A ∩ C) = 0.40 + 0.15 – 0 = 0.55 (P(X ∈ A ∩ C) = 0 since A and C are mutually exclusive) 315. a) P(X ≤ 7000) = 1 – P(X > 7000) = 1 – 0.45 = 0.55 b) P(X > 5000) = 1 – P(X ≤ 5000) = 1 – 0.05 = 0.95 c) P(5000 < X ≤ 7000) = P(X ≤ 7000) – P(X ≤ 5000) = 0.55 – 0.05 = 0.50 316. a) Probability that a component does not fail: ) E ( P 1 ′ = 1 – P(E 1 ) = 1 – 0.15 = 0.85 b) P(E 1 or E 2 ) = P(E 1 ) + P(E 2 ) = 0.15 + 0.30 = 0.45 c) P(E 1 or E 2 ) ′ = 1 – 0.45 = 0.55 317. a) P(X = 2 or X = 3) = 0.25 + 0.25 = 0.50 b) P(X < 2) = P(X = 0) + P(X = 1) = 0.10 +0.15 = 0.25 c) P(X > 3) = P(X = 4) + P(X = 5) = 0.15 + 0.10 = 0.25 d) P(X > 0) = 1 – P(X = 0) = 1 – 0.1 = 0.9d) P(X > 0) = 1 – P(X = 0) = 1 – 0....
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This note was uploaded on 12/11/2009 for the course CSE IEE taught by Professor Chattin during the Summer '09 term at ASU.
 Summer '09
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